Expand the following logarithmic expression:

log(16x²y²)^1/3

First off, we can pull that exponent out:

Now we can think of the stuff inside the log as 16x²y² = (4xy)², which means we can move that 2 to the front of the log and multiply it by the fraction that's already chilling there:

Split the log into three mini-logs using the sum of logs rule:

And that's it. We expanded everything as much as we possibly can.

Combine the following logarithmic expression into a single log:

3(log 4 + log a + log b)

Squeeze the three logs together using the sum of logs rule to kick off this crazy log-party:

3log(4ab)

Then use the exponent rule to sneak that 3 into the party. Don't forget to stick the exponent on the entire expression inside the log, not just the b:

log(4ab)³

Now simplify by cubing everything:

log(64a³b³)

Perfection.

Is y = log₂(4x) equivalent to 2^{(y – 2)} = x? If so, show how.

Let's run through a lightning round and see if we can turn the log equation into the exponential one.

Separate the logs on the right side:

y = log₂ 4 + log₂ x

Solve the left-hand log:

y = 2 + log₂ x

Isolate the term with x:

y – 2 = log₂ x

Exponentiate both sides with a base of 2:

2^{(y – 2)} = 2^{log₂ x}

Right-hand side exponent and log cancel:

2^{(y – 2)} = x

A perfect match! Yep, the two equations are equivalent.

Expand the following logarithmic expression:

First separate the numerator and denominator using the difference of logs rule:

log(2x²) – log(4x)

Nice. We can't pull that exponent out of the first log, though, because it's not being applied to the entire expression inside the log. If it were log(2x)², we could expand it to 2log(2x), but it's not, so we can't. Careful with those exponents.

Expand the following expression:

log(27x²)

First we separate our log into two terms:

log 27 + log x²

We can also rewrite 27 as 3³, which should let us pull that exponent:

log 3³ + log x²

3 log 3 + 2 log x

Expand the following log equation, and write it using only base-10 logarithmic functions. Solve irrationals to 3 decimal places:

y = ln[(4^e)(3x + 1)]

First, separate into two terms using the sum of logs rule:

y = ln 4^e + ln(3x + 1)

Use the change of base rule on both terms:

Solve irrationals: 

It ain't pretty, but it'll do the trick.

Is equivalent to 10^y = 81x²? If so, show how.

Let's mess with the first equation and see if we can get it to match the second. Ready?

First, multiply by 2 on both sides:

y = 2log(9x)

Use the property of exponents in logs to swing that 2 around:

y = log(9x)²

Distribute the exponent:

y = log(81x²)

Exponentiate both sides with a base of 10:

10^y = 81x²

Yussss. They match.

Can we simplify the equation y = (log x)(log 2) any further?

Nope! There aren't any logarithmic laws that tell us we can combine products of logs, and there's only one term in each log that can't be manipulated in any meaningful way. We can't whittle this guy down any further.